Question: What is the value of $\dfrac{d}{dx}\left(\dfrac{6}{x^4}+\dfrac{1}{x^2}\right)$ at $x=-1$ ?
The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=-1$. Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}\dfrac{6}{x^4}+\dfrac{1}{x^2} \\\\ &=6x^{-4}+x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(6x^{-4}+x^{-2} ) \\\\ &=6\dfrac{d}{dx}(x^{-4})+\dfrac{d}{dx}(x^{-2}) \\\\ &=6(-4x^{-5})+(-2)x^{-3} \\\\ &=-24x^{-5}-2x^{-3} \\\\ &=-\dfrac{24}{x^5}-\dfrac{2}{x^3} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{6}{x^4}+\dfrac{1}{x^2}\right)=-\dfrac{24}{x^5}-\dfrac{2}{x^3}$. Let's evaluate it at $x=-1$ : $\begin{aligned} &\phantom{=}-\dfrac{24}{x^5}-\dfrac{2}{x^3} \\\\ &=-\dfrac{24}{(-1)^5}-\dfrac{2}{(-1)^3} \\\\ &=24+2 \\\\ &=26 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{6}{x^4}+\dfrac{1}{x^2}\right)$ at $x=-1$ is $26$.